Unit 2: Exponential and Logarithmic Functions

Exponential Functions

Definition and Characteristics of Exponential Functions

• An exponential function is a function where the independent variable appears in the exponent, typically written as \(f(x) = a \cdot b^x\), where \(a \neq 0\), \(b > 0\), and \(b \neq 1\). The value of \(a\) is the initial amount, and \(b\) is the base that determines the growth or decay rate. The restriction \(b > 0\) and \(b \neq 1\) ensures the function is meaningful and non-constant.
• If \(b > 1\), the function represents exponential growth; if \(0 < b < 1\), it represents exponential decay. In growth, outputs increase rapidly over time; in decay, they decrease quickly at first and then level off. This rate of change is proportional to the function's current value.
• Exponential functions are continuous and smooth for all real numbers. Their domain is \((-\infty, \infty)\), and their range is \((0, \infty)\) when \(a > 0\). The horizontal asymptote is \(y = 0\) unless vertical shifts change the baseline.
• These functions model real-world processes where change happens by multiplication rather than addition. Examples include population growth, radioactive decay, and compound interest. Recognizing when data grows by equal factors over equal intervals helps identify exponential behavior.
• On a graph, exponential functions grow slowly at first, then more rapidly if \(b > 1\), or decrease rapidly at first, then slow down if \(0 < b < 1\). The y-intercept occurs at \((0, a)\) because \(b^0 = 1\).

Relating Geometric Sequences and Exponential Functions

• A geometric sequence is a sequence where each term is found by multiplying the previous term by a constant ratio \(r\). Its nth term is given by \(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term. This is structurally similar to the formula for an exponential function.
• If we treat the term number \(n\) as the independent variable and the term value \(a_n\) as the dependent variable, a geometric sequence can be seen as discrete points from an exponential function. The base \(b\) in the exponential function corresponds to the common ratio \(r\) in the sequence.
• For example, the sequence \(2, 6, 18, 54, \dots\) has \(a_1 = 2\) and \(r = 3\), so \(a_n = 2 \cdot 3^{n-1}\). As a function, this becomes \(f(x) = 2 \cdot 3^{x-1}\) for integer \(x\), which is an exponential growth function.
• Understanding this relationship allows you to model discrete repeated multiplication (like interest compounded yearly) using the same principles as continuous exponential growth. The only difference is whether the input variable is restricted to integers or allowed to vary continuously.
• In applied problems, recognizing when a pattern of values forms a geometric sequence can help you write the corresponding exponential function. This is especially useful in financial math, population studies, and computer science algorithms.

Interpreting Parameters in Context

• In \(f(x) = a \cdot b^x\), \(a\) represents the initial value when \(x = 0\). For example, if a city’s population starts at 50,000, \(a = 50,000\). This initial value anchors the function to a real-world starting point.
• The base \(b\) determines the growth or decay factor. If \(b > 1\), it grows by a factor of \(b\) each unit of \(x\); if \(0 < b < 1\), it decays by a factor of \(b\). For instance, \(b = 1.08\) means an 8% growth rate per time unit, while \(b = 0.92\) means an 8% decay rate.
• When expressing growth or decay rates as percentages, \(b = 1 + r\) for growth and \(b = 1 - r\) for decay, where \(r\) is the rate in decimal form. This makes it easy to switch between base form and rate form in applied problems.
• If the equation is written as \(f(x) = a \cdot b^{x-h} + k\), the parameter \(h\) shifts the graph horizontally, and \(k\) shifts it vertically. In context, \(k\) might represent a baseline level that the function approaches but never crosses, such as a limiting temperature or population cap.
• Interpreting parameters in context ensures your model matches the real-world situation. For example, in radioactive decay, \(a\) is the initial amount of substance, \(b\) is related to the half-life, and \(k = 0\) because the substance approaches zero over time.

Exponential Functions (Continued)

Transformations of Exponential Functions

• Transformations shift, stretch, or reflect the graph of an exponential function without changing its basic growth or decay nature. The general form \(f(x) = a \cdot b^{x-h} + k\) includes horizontal shifts (\(h\)), vertical shifts (\(k\)), and reflections/stretches controlled by \(a\). Understanding these helps match equations to graphs quickly.
• A horizontal shift occurs when \(h\) changes: \(f(x) = a \cdot b^{x-h} + k\) moves right if \(h > 0\) and left if \(h < 0\). For example, \(2^{x-3}\) is shifted 3 units to the right compared to \(2^x\). Always remember the shift direction is opposite the sign inside the exponent.
• A vertical shift occurs when \(k\) changes, moving the entire graph up (\(k > 0\)) or down (\(k < 0\)). This changes the horizontal asymptote from \(y = 0\) to \(y = k\), altering the baseline the function approaches as \(x \to \pm\infty\).
• Reflections occur when \(a\) is negative (reflects over the horizontal asymptote) or when the exponent’s coefficient is negative (reflects over the y-axis). These flips can drastically change the graph’s appearance while keeping the same general shape.
• Vertical stretches and compressions are controlled by \(|a|\). If \(|a| > 1\), the graph grows/decays more steeply; if \(0 < |a| < 1\), the graph is less steep. This affects how quickly the function moves away from its asymptote.

Solving Exponential Equations

• When the equation’s exponential terms have the same base, you can solve by setting the exponents equal. For example, \(3^{2x} = 3^5\) leads to \(2x = 5\), so \(x = 2.5\). This is the fastest method when matching bases is possible.
• If bases differ but can be rewritten to match, rewrite them before equating exponents. For instance, \(4^{x+1} = 8^{2x}\) becomes \((2^2)^{x+1} = (2^3)^{2x}\), then solve \(2x + 2 = 6x\) to find \(x = 0.5\).
• When bases cannot be rewritten to match, apply logarithms to both sides. For example, \(5^{x} = 14\) becomes \(\log(5^{x}) = \log(14)\), which simplifies to \(x \log 5 = \log 14\), giving \(x = \frac{\log 14}{\log 5}\).
• Natural logs (\(\ln\)) are especially useful when the base is \(e\), as they simplify directly: \(e^{3x} = 50\) → \(3x = \ln 50\) → \(x = \frac{\ln 50}{3}\). This is common in continuous growth/decay problems.
• Always check your solution in the original equation, especially if you’ve applied logarithmic rules, to ensure no extraneous results. This is particularly important if the equation came from a real-world context where negative or zero outputs are impossible.

Modeling Data Sets with Exponential Functions

• Exponential models are used when a data set changes by equal factors over equal intervals, not by equal differences. This is determined by checking if \(\frac{y_{n+1}}{y_n}\) is roughly constant for the data points. If so, an exponential function is appropriate.
• To find the model, first determine \(a\), the starting value, from the y-intercept or the first data point. Then, calculate \(b\) (the growth/decay factor) as \(b = \left(\frac{y_{\text{final}}}{y_{\text{initial}}}\right)^{1/n}\), where \(n\) is the number of intervals between those points.
• Once \(a\) and \(b\) are known, the model can be written as \(y = a \cdot b^x\). This model should then be tested by substituting other data points to check accuracy. If the fit is poor, you may need a regression method.
• Most graphing calculators and software have exponential regression features that compute \(a\) and \(b\) directly from data. In TI calculators, this is often labeled “ExpReg” under the STAT → CALC menu. This is especially useful for large data sets.
• Always interpret the parameters in context. For example, in \(y = 200 \cdot 0.85^x\), \(200\) might be the initial number of customers, and \(0.85\) means the number decreases by 15% each day. Contextual interpretation is crucial for explaining your results.

Modeling with Exponential Functions

Growth and Decay Models

• Exponential growth occurs when a quantity increases by the same percentage over equal time intervals. The general model is \(y = a \cdot (1 + r)^t\), where \(a\) is the initial value, \(r\) is the growth rate in decimal form, and \(t\) is time. For example, a population growing 4% per year is modeled by \(P(t) = P_0 \cdot 1.04^t\).
• Exponential decay happens when a quantity decreases by the same percentage over equal time intervals. The decay model is \(y = a \cdot (1 - r)^t\), with \(r\) as the decay rate in decimal form. Radioactive decay is a classic example, where the amount of substance reduces by a constant percentage each half-life.
• For continuous growth or decay, the model \(y = a \cdot e^{kt}\) is used, where \(k > 0\) for growth and \(k < 0\) for decay. This formula is common in biology, physics, and finance for processes that change continuously rather than in discrete steps.
• In compound interest, the model \(A = P\left(1 + \frac{r}{n}\right)^{nt}\) calculates the balance \(A\) after \(t\) years, with principal \(P\), annual interest rate \(r\), and \(n\) compounding periods per year. For continuous compounding, the formula becomes \(A = Pe^{rt}\).
• Always check if exponential behavior is reasonable for the real-world situation. For instance, population growth cannot continue exponentially forever due to resource limits, so the model might only be valid for short-term predictions.

Logarithmic Functions

Definition and Characteristics of Logarithmic Functions

• A logarithmic function is the inverse of an exponential function and is written as \(y = \log_b(x)\), where \(b > 0\) and \(b \neq 1\). It answers the question: "To what power must \(b\) be raised to get \(x\)?" For example, \(\log_2(8) = 3\) because \(2^3 = 8\).
• The domain of a logarithmic function is \(x > 0\) because the argument of a log must be positive. Its range is all real numbers, and it has a vertical asymptote at \(x = 0\) for the basic parent function.
• Logarithmic functions are continuous and smooth on their domain but increase very slowly as \(x\) gets large. This makes them ideal for modeling processes where growth slows over time, such as the pH scale or the Richter scale.
• The base \(b\) affects the graph’s steepness. If \(b > 1\), the graph increases as \(x\) increases; if \(0 < b < 1\), the graph decreases as \(x\) increases. This is the opposite of exponential functions in terms of growth direction.
• Every logarithmic function can be rewritten in exponential form: \(\log_b(x) = y\) is equivalent to \(b^y = x\). This property is crucial for solving equations and understanding the log-exponential relationship.

Common Logarithmic Bases

Base 10 (Common Logarithm) and Base \(e\) (Natural Logarithm)

• The common logarithm, \(\log(x)\), has base 10 and is often used in scientific notation, pH, and decibel calculations. It’s widely supported on calculators without needing to specify the base. For example, \(\log(1000) = 3\) because \(10^3 = 1000\).
• The natural logarithm, \(\ln(x)\), has base \(e \approx 2.718\) and is frequently used in calculus, continuous growth/decay problems, and advanced science applications. For example, \(\ln(e^4) = 4\) because \(e^4\) equals \(e\) raised to the fourth power.
• Changing between bases can be done using the change of base formula: \(\log_b(x) = \frac{\log_k(x)}{\log_k(b)}\), where \(k\) can be any positive base. This is especially helpful when your calculator only supports \(\log\) and \(\ln\).
• Common and natural logs appear in many real-world contexts because they align with our measurement systems. Base 10 logs fit well with metric scaling, while base \(e\) logs naturally arise from continuous processes.
• In practical applications, choosing the correct base depends on the context: use base \(e\) for continuous modeling and calculus, and base 10 for orders of magnitude and scaling measurements.

Logarithmic Functions (Continued)

Properties of Logarithms

• Product Rule: \(\log_b(M \cdot N) = \log_b M + \log_b N\). This means the log of a product is the sum of the logs of the factors. For example, \(\log_2(8 \cdot 4) = \log_2(8) + \log_2(4) = 3 + 2 = 5\). This is useful for breaking down multiplication inside logs into simpler terms.
• Quotient Rule: \(\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N\). The log of a quotient is the difference of the logs of numerator and denominator. For example, \(\log_5\left(\frac{125}{5}\right) = \log_5(125) - \log_5(5) = 3 - 1 = 2\).
• Power Rule: \(\log_b(M^p) = p \cdot \log_b M\). Exponents inside a log can be moved out front as multipliers. For example, \(\log_3(81) = \log_3(3^4) = 4 \log_3(3) = 4\). This is especially useful when solving equations.
• Change of Base Formula: \(\log_b(M) = \frac{\log_k M}{\log_k b}\), where \(k\) is any positive base. This allows you to evaluate logs with any base on calculators that only have \(\log\) (base 10) and \(\ln\) (base \(e\)).
• These properties work for all positive \(M\) and \(N\) where \(b > 0\) and \(b \neq 1\). They are often combined in multi-step simplifications, such as expanding \(\log_2\left(\frac{8x^3}{y^2}\right)\) into \(\log_2(8) + 3\log_2(x) - 2\log_2(y)\).

Solving Logarithmic Equations

• If a log equation is in the form \(\log_b(x) = y\), rewrite it in exponential form: \(x = b^y\). For example, \(\log_4(x) = 3\) becomes \(x = 4^3 = 64\). This conversion is the most direct method when a single log is isolated.
• When there are multiple logs, use the properties of logarithms to combine them into a single log before rewriting in exponential form. For instance, \(\log(x) + \log(x-3) = 2\) becomes \(\log[x(x-3)] = 2\), then rewrite as \(x(x-3) = 10^2\).
• Always check for extraneous solutions after solving, because taking logs can introduce invalid results. For example, solving \(\log(x-2) = 1\) gives \(x = 12\), which is valid, but if the result made the log’s argument negative, it would have to be discarded.
• If the equation involves different log bases, either convert all logs to the same base using the change of base formula or rewrite into exponential form using the given base. Consistency is key in multi-step problems.
• In applied problems, the final step often involves interpreting the solution in context, such as determining the time for a population to reach a certain size or the pH of a solution given hydrogen ion concentration.

Graphing Logarithmic Functions

• The parent graph of \(y = \log_b(x)\) has a vertical asymptote at \(x = 0\) and passes through the point \((1, 0)\). It increases if \(b > 1\) and decreases if \(0 < b < 1\). The graph only exists for \(x > 0\) because logs are undefined for zero or negative inputs.
• Transformations follow similar rules to exponential functions: \(y = \log_b(x - h) + k\) shifts the graph horizontally by \(h\) units and vertically by \(k\) units. Reflections and stretches occur with negative or large coefficients.
• When graphing by hand, it’s helpful to choose x-values that are powers of the base so that the log values are integers. For example, for \(\log_2(x)\), choose \(x = 1, 2, 4, 8\) to get \(y = 0, 1, 2, 3\).
• The range of a log function is all real numbers, but the domain is strictly \(x > 0\). Any transformation that shifts the graph horizontally changes the domain restriction accordingly.
• In modeling contexts, the vertical asymptote represents a boundary beyond which the situation is undefined — for example, the minimum time after which a chemical reaction’s measurement becomes meaningful.

Modeling Scenarios with Logarithmic Functions

Identifying Logarithmic Models

• Logarithmic models are used when a quantity increases or decreases rapidly at first and then levels off over time. This is common in learning curves, cooling processes, and some growth phenomena where the rate of change decreases as the value increases.
• The general form is \(y = a + b \cdot \log_b(x)\) or \(y = a + b \cdot \ln(x)\), where \(a\) represents a baseline value and \(b\) determines the rate of change. The choice of base often depends on the context; \(\ln\) is common for continuous-time processes.
• Examples include the pH scale (\(\text{pH} = -\log[H^+]\)), earthquake magnitude (\(M = \log\left(\frac{A}{A_0}\right)\)), and sound intensity in decibels (\(dB = 10 \log\left(\frac{I}{I_0}\right)\)). Each uses a logarithm to compress a wide range of values into a manageable scale.
• To create a logarithmic model from data, first check if the rate of change decreases as the input increases. Plotting the data can reveal if the curve resembles a log shape. Regression tools can then be used to fit a log model to the data.
• In applied contexts, interpret parameters carefully. For example, in \(y = 50 + 20\ln(x)\), 50 might represent an initial level, and 20 controls how steeply the quantity increases at small \(x\)-values before leveling off.

Relationship Between Exponential and Logarithmic Functions

Understanding the Inverse Relationship

• Exponential and logarithmic functions are mathematical inverses. If \(y = b^x\), then \(x = \log_b(y)\). This means applying an exponential function and then a log with the same base (or vice versa) returns the original value.
• Their graphs are reflections of each other across the line \(y = x\). This symmetry is useful when graphing one function if you already know the other. For example, knowing the graph of \(y = 2^x\) makes it easy to sketch \(y = \log_2(x)\).
• Because they are inverses, the domain of an exponential function is the range of its logarithmic inverse, and vice versa. For \(y = b^x\), the domain is all real numbers and the range is \(y > 0\); for \(y = \log_b(x)\), the domain is \(x > 0\) and the range is all real numbers.
• This inverse relationship is the foundation for solving exponential equations with logarithms and logarithmic equations with exponentials. For instance, to solve \(3^x = 20\), take the log base 3 of both sides to isolate \(x\).
• In real-world modeling, this connection allows for switching between multiplicative processes (exponential) and processes that "undo" multiplication (logarithmic), such as converting between population size and time or between pH and hydrogen ion concentration.

Composing Functions and Finding Inverses

Function Composition

• Composition combines two functions so the output of one becomes the input of the other. For example, if \(f(x) = 2^x\) and \(g(x) = x + 3\), then \(f(g(x)) = 2^{x+3}\). This is common in modeling scenarios where multiple transformations occur sequentially.
• Order matters in composition: \(f(g(x))\) is generally different from \(g(f(x))\). For example, \(g(f(x)) = 2^x + 3\) is not the same as \(2^{x+3}\). Always evaluate the inner function first.
• Compositions of exponentials and logarithms often simplify due to their inverse nature. For instance, \(\log_b(b^x) = x\) and \(b^{\log_b(x)} = x\) for \(x > 0\). Recognizing these simplifies complex expressions quickly.
• In applications, composition is used to represent processes that depend on multiple steps, such as a quantity that grows exponentially and is then scaled or shifted before observation.
• Graphically, composition can be interpreted as applying one transformation after another, which affects the domain and range according to both functions involved.

Finding Inverse Functions

• To find the inverse of a function, swap \(x\) and \(y\) and solve for the new \(y\). For example, if \(y = 3^x\), swapping gives \(x = 3^y\); solving for \(y\) yields \(y = \log_3(x)\), the inverse function.
• For logarithmic functions, the process is the same: if \(y = \log_b(x)\), swapping gives \(x = \log_b(y)\); rewriting in exponential form gives \(b^x = y\). This results in an exponential inverse.
• Always check that the composition of a function and its inverse returns the original input: \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\). This confirms correctness.
• When finding inverses involving shifts and stretches, apply the inverse transformations in reverse order. For instance, if \(f(x) = 2^{x-4} + 5\), subtract 5, take the log base 2, then add 4 when finding the inverse.
• In modeling, inverses allow you to solve for input variables given output measurements — for example, finding the time it takes for an investment to double by inverting the growth function.

Validating a Function Model Using a Residual Plot

Understanding Residuals

• A residual is the difference between the observed value and the predicted value from a model: \(\text{Residual} = \text{Observed} - \text{Predicted}\). Residuals tell us how far off our model’s predictions are from actual data points.
• Residuals are positive if the model underestimates the data point and negative if the model overestimates. Analyzing the sign and size of residuals helps detect consistent bias in the model.
• Residuals close to zero indicate that the model’s prediction is accurate for that point. Large residuals indicate the model is not capturing the data’s behavior in that region.
• Residuals are measured in the same units as the dependent variable, so interpretation should be made in the context of the real-world situation being modeled.
• In exponential and logarithmic models, residuals often highlight whether the chosen model type fits the data pattern or if a different functional form is needed.

Interpreting a Residual Plot

• A residual plot is a graph of residuals versus the independent variable. It helps visually assess whether a model is appropriate for the data.
• If residuals are randomly scattered around zero without any clear pattern, the model is likely a good fit. This indicates the model’s errors are random and not systematic.
• If the residual plot shows a pattern (like a curve, trend, or clusters), the model may be missing an important feature of the relationship. This means a different model might be more appropriate.
• Residual plots can also indicate heteroscedasticity, where the spread of residuals changes with the independent variable. This suggests that model accuracy varies across the range of data.
• For exponential and logarithmic models, a systematic curve in the residual plot often means a polynomial or power model might better capture the data’s pattern.

Limitations of Exponential and Logarithmic Models

Recognizing Model Boundaries

• Exponential models assume constant percent change over equal intervals, which is often unrealistic for long-term predictions. For example, populations cannot grow exponentially forever due to resource limits.
• Logarithmic models assume rapid change initially followed by a slowing rate of change. They are not suitable if the quantity continues to grow or decline at a steady rate indefinitely.
• Both models assume that the relationship between variables is smooth and continuous. If real-world data have abrupt jumps or irregular patterns, these models may fail.
• Exponential decay models cannot account for situations where the quantity drops below zero or becomes negative, as this violates the model’s mathematical structure.
• Logarithmic models cannot handle zero or negative inputs because the logarithm is undefined for these values. This limits their applicability to certain domains.

Impact of Assumptions on Accuracy

• If the percent change in an exponential model changes over time, the model will lose accuracy quickly. This is common in financial markets, where growth rates fluctuate.
• For logarithmic models, if the early rapid change doesn’t match the actual data trend, the model will either overpredict or underpredict significantly in the beginning.
• External factors not included in the model can drastically affect predictions. For example, sudden technological advances can cause exponential growth to speed up beyond what was modeled.
• Misidentifying whether a process is exponential, logarithmic, or neither leads to poor fits and misleading conclusions. Always test multiple model types against the data.
• Awareness of these limitations ensures that models are applied in contexts where their assumptions hold, increasing predictive reliability.

Common Misconceptions

Exponential Functions

• Misconception: Exponential growth always means rapid change. In reality, small growth rates (e.g., 1% per year) produce slow changes at first, with speed increasing only over time.
• Misconception: Exponential functions can have negative outputs. When \(a > 0\), exponential functions are always positive, and only vertical shifts can move them below zero.
• Misconception: The base \(b\) in \(a \cdot b^x\) can be negative. Negative bases cause undefined values for non-integer exponents, so \(b\) must be positive and not equal to 1.
• Misconception: Doubling time is the same for all exponential functions. Doubling time depends on the growth rate, and even small changes in \(r\) drastically affect it.
• Misconception: Exponential decay reaches zero. It approaches zero but never actually reaches it, due to the horizontal asymptote.

Logarithmic Functions

• Misconception: Logarithmic functions can take zero or negative arguments. Logs are only defined for positive inputs.
• Misconception: Logarithmic growth is the same as linear growth. Logarithmic growth slows over time and eventually becomes very small, unlike steady linear change.
• Misconception: The base of a log doesn’t matter. Changing the base changes the steepness and scale of the graph, which can alter model interpretation.
• Misconception: Logs are only useful in math class. In reality, they are essential for modeling real-world scales like pH, earthquake magnitude, and sound intensity.
• Misconception: A vertical asymptote at \(x=0\) means the graph never increases. While the graph increases slowly, it continues without bound as \(x\) increases.

Practice Problem 1: Exponential Modeling & Interpretation

Problem

• A culture of bacteria is measured every hour. The recorded counts (in thousands) are: \(t=0\): 6.0, \(t=2\): 9.72, \(t=5\): 17.80.
  (a) Use the first two data points to build an exponential model \(N(t)=a\cdot b^{t}\).
  (b) Interpret \(a\) and \(b\) in context and convert \(b\) to a percent growth rate per hour.
  (c) Use your model to predict \(N(5)\) and compare to the observed value by computing a residual.
  (d) Find the doubling time using your model.
  (e) State one limitation/assumption of using this exponential model for long-term prediction.

Solution

• (a) Build the model. From \(t=0\), \(N(0)=6.0\Rightarrow a=6.0\) because \(b^{0}=1\). Using \(t=2\), \(N(2)=9.72=6.0\cdot b^{2}\Rightarrow b^{2}=\frac{9.72}{6.0}=1.62\). Thus \(b=\sqrt{1.62}\approx1.2730\) and the model is \( \displaystyle N(t)=6.0\cdot (1.2730)^{t}\).
• (b) Interpret parameters. The parameter \(a=6.0\) (thousand) is the initial population at \(t=0\), so the culture starts at \(6000\) bacteria. The base \(b\approx1.2730\) means each hour the population is multiplied by about \(1.2730\). As a percent rate, \(r=b-1\approx0.2730\), so the culture grows about \(27.30\%\) per hour.
• (c) Predict and compute residual. Model prediction at \(t=5\) is \( \hat N(5)=6.0\cdot (1.2730)^{5}\). Compute \((1.2730)^{5}\approx 3.279\), giving \( \hat N(5)\approx 6.0\cdot 3.279=19.674\) (thousand). The observed is \(17.80\) (thousand), so the residual is \(\text{Obs}-\text{Pred}=17.80-19.674\approx -1.874\) (thousand), indicating the model overestimates at \(t=5\).
• (d) Doubling time. For \(N(t)=a b^{t}\), doubling time \(T_d\) satisfies \(b^{T_d}=2\Rightarrow T_d=\dfrac{\ln 2}{\ln b}\). With \(b\approx1.2730\), \(T_d\approx \dfrac{\ln 2}{\ln(1.2730)}\approx \dfrac{0.6931}{0.2410}\approx 2.875\) hours. Interpretation: under the model, the culture doubles roughly every \(2.88\) hours.
• (e) Limitation/assumption. The model assumes a constant percent growth rate and unlimited resources, which is rarely true over long intervals. As nutrients deplete or waste accumulates, growth slows, so extrapolations far beyond the observed window may substantially overpredict actual counts.

Practice Problem 2: Solving with Logs, Inverses, and Half-Life

Problem

• A radioactive isotope decays according to \(A(t)=A_0 e^{kt}\) with \(k<0\).
  (a) If the half-life is \(h=12.3\) years, find \(k\).
  (b) A sample starts at \(A_0=250\) mg. How much remains after \(t=30\) years?
  (c) Write the inverse function \(t(A)\) that solves for time given a remaining amount \(A\).
  (d) Use your inverse to find how long until the sample reaches \(60\) mg.
  (e) Explain why logarithms are the appropriate tool in parts (a), (c), and (d).

Solution

• (a) Find \(k\) from half-life. Half-life means \(A(h)=\frac{A_0}{2}=A_0 e^{kh}\). Divide by \(A_0\): \(\frac{1}{2}=e^{kh}\Rightarrow \ln\!\left(\frac{1}{2}\right)=kh\). Thus \(k=\dfrac{\ln(1/2)}{h}=\dfrac{-\ln 2}{12.3}\approx \dfrac{-0.6931}{12.3}\approx -0.05633\ \text{yr}^{-1}\). This negative rate encodes exponential decay per year.
• (b) Amount after 30 years. Use \(A(t)=250\,e^{kt}\) with \(k\approx -0.05633\) and \(t=30\): \(A(30)=250\,e^{-0.05633\cdot 30}\). Compute the exponent \(-0.05633\cdot 30\approx -1.6899\), so \(A(30)\approx 250\cdot e^{-1.6899}\approx 250\cdot 0.1846\approx 46.2\ \text{mg}\). The sample has fallen well below one half-life by 30 years.
• (c) Inverse function \(t(A)\). Start with \(A=A_0 e^{kt}\). Divide to isolate the exponential: \(\frac{A}{A_0}=e^{kt}\). Take natural logs: \(\ln\!\left(\frac{A}{A_0}\right)=kt\). Solve for time: \( \displaystyle t(A)=\frac{1}{k}\,\ln\!\left(\frac{A}{A_0}\right)\). This inverse lets us compute the time needed to reach any positive amount \(A\).
• (d) Time to reach 60 mg. Plug \(A=60\), \(A_0=250\), and \(k\approx -0.05633\) into \(t(A)\): \( \displaystyle t=\frac{1}{-0.05633}\ln\!\left(\frac{60}{250}\right)\). Compute the ratio \(60/250=0.24\) and \(\ln(0.24)\approx -1.4271\). Then \(t\approx \frac{-1.4271}{-0.05633}\approx 25.33\ \text{years}\), so it takes about \(25.3\) years to decay to \(60\) mg.
• (e) Why logs? Logarithms undo exponentials, converting multiplicative growth/decay into additive linear terms that are solvable for \(t\) or \(k\). In (a), taking \(\ln\) isolates \(k\) from the half-life definition; in (c)–(d), \(\ln\) transforms \(e^{kt}\) so we can solve directly for time. Without logs, the exponent \(t\) remains trapped and cannot be isolated algebraically.